Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $F$ be a finite field of characteristic $2$. Let $m \geq 2 $ be a positive integer. Seems unknown if the trinomial $$ T(t,x) = x^{2m+1}+x^2+s(t) \in F[t][x] $$ (more explicitly, the constant coefficient $s(t)$ is a polynomial in $t$, i.e., $s(t) \in F[t]$)

has monic factors $D(t,x)$ in $F[t][x]$ with degree $3$ relative to $x$.

I.e., $$ D(t,x) = x^3+ a_2(t)x^2+a_1(t)x + a_0(t) $$ with $a_2(t), a_1(t),a_0(t) \in F[t]$ and $T(t,x) = D(t,x)K(t,x)$ for some $K(t,x) \in F[t][x].$

We will then say that $T(t,x)$ have $D(t,x)$ as a factor or that $D(t,x)$ divides $T(t,x)$.

Question: Assume that $D(t,x)$ as above divides the trinomial $T(t,x).$

Do we have $$ \deg(a_1(t)) = 2 \deg(a_2(t)),\;\;\deg(a_0(t))=3\deg(a_2(t)). $$ ???

I am aware of the work of Schinzel on trinomials. The trinomials $T(t,x)$ do not seem to be worked out in these papers.

share|improve this question
    
Is $s(t)$ an arbitrary element of $F(t)$? –  S. Carnahan May 24 '11 at 4:56
    
In line 3, we take $s(t)$ as an arbitrary element of $F[t]$. You think is more useful to take the $s(t)$ in the full ring of quotients $F(t)$ ? –  Luis H Gallardo May 24 '11 at 7:53
    
Sorry, that was a typo. I meant square brackets. –  S. Carnahan May 24 '11 at 10:08
add comment

2 Answers

up vote 4 down vote accepted

We show the following:

Theorem. Let $F$ be a field of characteristic $2$, and $s(t)\in F[t]$ a non-constant polynomial. Then $f(t,X)=X^{2m+1}+X^2+s(t)$ (where $m\in\mathbb N$) is either irreducible, or a product of irreducible factors of degrees $1$ and $2m$.

In particular, this answers the question about the non-existence of cubic factors for $m\ge2$.

We use Galois theory, group theory and a little bit of valuation theory: Set $y=s(t)$, and let $x$ be a root of an irreducible factor of degree $k$ of $f(t,X)$. Note that $x^{2m+1}+x^2=s(t)=y$. We obtain the following field degrees: $[F(x):F(y)]=2m+1$ and $[F(x,t):F(t)]=k$.

It is easy to see that we may assume that $s(t)$ is not a polynomial in $t^2$. (Take the derivative with respect to $t$ of a factorization). So $x$ and $t$ are separable over $F(y)$. Let $L$ be the smallest Galois extension of $F(y)$ which contains $x$ and $t$, and set $G=\text{Gal}(L/K(y))$.

The group $G$ need not act faithfully on the conjugates of $x$. Let $N$ be the kernel of this action. So $G/N$ acts faithfully on the conjugates of $x$. Note that $G/N$ is the Galois group of $X^{2m+1}+X^2+y$ over $F(y)$.

We first claim that $G/N$ is the symmetric group $S_{2m+1}$: The polynomial $X^{2m+1}+X^2$ is functionally indecomposable. To see this, take the derivative of $X^{2m+1}+X^2=A(B(X))$. A short calculation shows that either $A$ or $B$ is linear. So, by Luroth's Theorem, $G/N$ is primitive. Furthermore, $X^{2m+1}+X^2=(X^{2m-1}-1)X^2$, with the first factor being separable. Valuation theory shows that the inertia generator of a place of $L$ lying above the place $y\mapsto 0$ is a transposition on the conjugates of $x$. A well-known theorem of group theory then shows that $G/N$ acts as the symmetric group $S_{2m+1}$ on the conjugates of $x$. Let $G_x$ and $G_t$ be the stabilizers of $x$ and $t$ in $G$. As $x$ has degree $k$ over $F(t)$, we get that $G_t$ has an orbit of length $k$ on the conjugates of $x$. But $NG_t$ has the same orbits as $G_t$ on this set. Thus, we may (and do) replace $G_t$ with $NG_t$, which amounts to replacing $s(t)$ with a polynomial $s_1(t)$ such that $s(t)=s_1(s_2(t))$ for another polynomial $s_2(t)$, and $X^{2m+1}+X^2+s_1(t)$ has the same factorization pattern as $f(t,X)$.

So we have $N\le G_t$, that is $N$ fixes $t$. But $N$ is normal in $G$, so $N$ fixes all conjugates of $t$ (and those of $x$ by definition), hence $N=1$.

So $G=S_{2m+1}$ acts faithfully on the conjugates of $x$.

Now let $T$ be the inertia group of a place of $L$ lying above the place $p=(y\mapsto\infty)$. As $p$ is totally ramified in $F(x)$, and $F(x)/F(y)$ is tame, we get that $T$ is a cyclic group permuting regularly the conjugates of $x$. In particular, $T$ has order $2m+1$. However, $p$ is totally ramified in $K(t)$ too, so $T$ permutes the conjugates of $t$ transitively as well. Therefore $[G:G_t]\le 2m+1$.

Now suppose that $2\le k\le 2m-1$, so $G_t$ has an orbit of length $k$ on the conjugates of $x$. This yields $\lvert G_t\rvert\le k!(2m+1-k)!$, hence $[G:G_t]\ge\binom{2m+1}{k}>2m+1$, contrary to the inequality above.

If $k=1$, then $G_x$ and $G_t$ are conjugate in $G$. In particular $G_t$ has orbit lengths $1$ and $2m$ on the conjugates of $x$, and the claim follows again.

share|improve this answer
    
Very nice argument! –  Igor Rivin Nov 22 '12 at 18:00
add comment

Sorry, this is only worth a comment, but I'm a new guy here, so not enough rep to do that :-)

As you discuss trinomials, the silly counterexample of $(x^3+x^2+x)(x^2+x)=x^5+x^2$ is probably not interesting. Anyway, for which values of $m$ have you verified this? Here's how it goes in the case $m=2$ (you undoubtedly know this, but let's get the ball rolling).

So assume $m=2$. In order to make the quartic and cubic terms vanish from the product, we must set $K(x)=x^2+a_2(t) x+ [a_1(t)+a_2(t)^2]$, so we can compute the product (I leave out the $t$:s for now): $$ D(x)K(x)=x^5+x^2(a_0+a_2^3)+x(a_0a_2+a_1a_2^2+a_1^2)+(a_0a_1+a_0a_2^2)=T(x). $$ A comparison of quadratic terms gives us the relation $a_0=1+a_2^3$. This gives your second claimed relation unless $a_2(t)=1$ (, which leads to the silly case above). Then a comparison of the linear terms gives the relation $$ 0=a_2+a_2^4+a_1a_2^2+a_1^2. $$ If here $\deg a_1>2\deg a_2$, then the last term has a higher degree than the others, which is no-no (non-archimedean triangle inequality w.r.t to the infinite place). OTOH if $\deg a_1<2\deg a_2$, then the term $a_2^4$ dominates the others, and we again have a contradiction.

This is admittedly a very elementary approach, but do you know how far you can go with methods like this? We can always determine the coefficients of the other factor by using the known coefficients (down to the cubic term) of the product, and then get conditions by comparing the quadratic and linear terms. I am uncertain as to how many values of $m$ I want to check by hand :-)

share|improve this answer
    
Thanks for discussing these case. –  Luis H Gallardo Jun 5 '11 at 17:19
    
@Luis: Was an example like the case $m=2$ above the basis of your conjecture/question? –  Jyrki Lahtonen Jun 6 '11 at 9:22
    
@Jyrki: Not indeed, instead the study of the possible degree $2$ factors of the trinomial. See my paper with Berrondo on the question. –  Luis H Gallardo Jun 6 '11 at 11:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.